Thread: differentiable function of 3 variables

1. differentiable function of 3 variables

If someone can help, please do help me for this question. I got stuck with it since yesterdays.

Prove that if f is a differentiable function of 3 variables and g(x,y,z) = f(x-y, y-z, z-x), then gsubx(x,y,z) + gsuby(x,y,z) +gsubz(x,y,z)=0.

Hint Let u =x-y, v=y-z, w=z-x, and write g(x,y,z)=f(u(x,y,z), v(x,y,z), w(x,y,z))

Thank you very much.

2. Originally Posted by littlemu
If someone can help, please do help me for this question. I got stuck with it since yesterdays.

Prove that if f is a differentiable function of 3 variables and g(x,y,z) = f(x-y, y-z, z-x), then gsubx(x,y,z) + gsuby(x,y,z) +gsubz(x,y,z)=0.

Hint Let u =x-y, v=y-z, w=z-x, and write g(x,y,z)=f(u(x,y,z), v(x,y,z), w(x,y,z))

Thank you very much.
You have,
g(u,v,w) with u=x-y and v=y-z and w=z-x

Now you can use the Chain Rule.

,

if f(u v w) is differentiable and u=x-y v=y-z and w=z-x show that

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